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alexandr402 [8]
2 years ago
15

Renee wants to put a fence around her

Mathematics
1 answer:
kramer2 years ago
3 0

Answer:

perimeter = 322.5 ft ( to the nearest tenth of a foot)

Step-by-step explanation:

Length the length of the sides of the square be L

Area of a square = L²

∴ 6,500 = L²

L = √(6,500)

L = 80.6226

Next Let us calculate the perimeter

Perimeter of a square = L + L + L + L = 4L

perimeter = 4 × 80.6226

Perimeter = 322.4904 ft

perimeter = 322.5 ft ( to the nearest tenth of a foot)

<em>N:B rounding to the nearest tenth of a foot is same as rounding to 1 decimal place.</em>

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The point P(-5,-8) is reflected across the x-axis to create point Q.
Ivahew [28]

Answer:

Answer: D. Q(5, −6) and R(−5, 6)

Step-by-step explanation:

3 0
3 years ago
Can the sum of two mixed numbers be equal to 2, explain why or why not
Sergeeva-Olga [200]
No because a mixed number is a whole number with a fraction and the smallest whole number is one so 1 and a fraction pulse and a fraction will equal more than 2
4 0
3 years ago
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Distance between parallel lines y=3x+10 and y=3x-20
Alecsey [184]

1. Take an arbitrary point that lies on the first line y=3x+10. Let x=0, then y=10 and point has coordinates (0,10).


2. Use formula d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} to find the distance from point (x_0,y_0) to the line Ax+By+C=0.


The second line has equation y=3x-20, that is 3x-y-20=0. By the previous formula the distance from the point (0,10) to the line 3x-y-20=0 is:

d=\dfrac{|3\cdot 0-10-20|}{\sqrt{3^2+(-1)^2}}=\dfrac{30}{\sqrt{10}}=3\sqrt{10}.


3. Since lines y=3x+10 and y=3x-20 are parallel, then the distance between these lines are the same as the distance from an arbitrary point from the first line to the second line.


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4 0
2 years ago
Match these similar triangles proof<br>​
AlekseyPX

Answer:

B

C

F

D

H

A

G

E

Step-by-step explanation:

Ok. They are trying to reconstruct the smaller looking triangle in the bigger triangle using angle A as the common angle.

The first statement is always the given.

Second they constructed line segment XY into the bigger triangle so that XY is parallel to BC.

Third, from the construction of the parallel lines we can now find corresponding angles that are congruent. This would be the use of F.

Since we have all three angles in triangle AXY and triangle ABC, then the construction of the smaller triangle we made inside the bigger triangle is similar to the bigger triangle. So we have the triangles are similar. You could say E or D here in my opinion. This is choice D.

Fifth the creation of those fractions of sides being equal comes from us knowing the corresponding sides of similar triangles are proportional. This is choice H.

Things looked cut off for the sixth thing so I can't fully read it, but it is possible a substitution has occured.

The seventh thing is a congruence statement which can be proven by a congruence postulate. The only one listed is SAS. So that is G.

The last thing, since the triangle construction is congruent to the smaller triangle then we know the smaller triangle is also similar to the bigger triangle since the bigger one is also similar to the construction we made. I really think E and D is interchangeable. Choice E goes here.

7 0
3 years ago
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